Exercise 2.4

Read Exercise 2.4 ~ Solution

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When I first read about defining cons, car and cdr procedurally, I had my first lisp wow moment.

(define (cons x y)
  (lambda (m) (m x y)))

(define (car z)
  (z (lambda (p q) p)))

(define (cdr z)
  (z (lambda (p q) q)))

(define test-pair (cons 3 4))
; what kind of beast are we dealing with
test-pair
# 
(car test-pair)
3
(cdr test-pair)
4

(cdr (cons x y))
(z (lambda (p q) q))
z = (cons x y) = (lambda (m) (m x y))
-> ((lambda (m) (m x y)) (lambda (p q) q))
-> ((lambda (p q) q) x y)
-> y